)a computer has an 8 gbyte memory with 64 bit word sizes. each block of memory stores 16 words. the computer has a direct-mapped cache of 128 blocks. the computer uses word level addressing. what is the address format? if we change the cache to a 4- way set associa: ve cache, what is the new address format?

a. the address format can be computed in two format:

i.  for memory addresses

solution

8GB = 1GW (64 bit word = 8 bytes/word), which requires 30 bits (log 2 1G=30). The memory address is divided into line number and word number. With 16 words per line,

we need 4 bits for the line number leaving 26 bits of the line number: 26 –4.

ii. for catch address

refill lines = 128 blocks x 8 bits ( 1 byte) = 1024 direct-mapped cache refill lines

The direct-mapped cache has 1024 refill lines so requires 10 bits for the line number,

leaving 26 – 10 = 16 bits for the tag: 16 – 10 – 4

b.  The 4-way set associative cache would have 256 lines in each of 4 sets, so we need 8 bits

for the line number leaving 19 bits for the tag: 18 – 8 – 4.

Explanation:

ipr and related laws.